Suppose $M$ is a manifold. We call a map $F:M \rightarrow \mathbb{R}^m$ proper if for every compact subset $K \subset \mathbb{R}^m$ we have that $F^{-1}(K) \subset M$ is compact.
Suppose now that $F: M \times [0,T) \rightarrow \mathbb{R}^m$ is a smooth family of immersions such that $F(\cdot ,0)$ is proper. Is it true that $\forall t \in [0,T): F(\cdot, t)$ is proper?


This is false as stated. For an (analytic) counterexample, take $M = \mathbb{R}$ and define $F:\mathbb{R} \times [0,\pi] \to \mathbb{R}$ via

$$F(x,t):= (1-\sin(t))x + \sin(t) \arctan(x).$$

For any fixed $t\in [0,\pi]$, $F(\cdot,t)$ is an immersion because the maximum slope of $\arctan$ is $1$. Clearly $F(\cdot,0)$ is proper. But $F(\cdot,\pi/2) = \arctan$, which is bounded and therefore not proper.

The conjecture is still false if $m > \text{dim}(M)$. As a counterexample, take $M= \mathbb{R}$ again, let $F$ be as above and define the map $G: \mathbb{R}\times [0,\pi]\to\mathbb{R}^m$ via

$$ G(x,t):= (F(x,t),\ldots,F(x,t)).$$

$G(\cdot,t)$ is an immersion for any $t \in [0,\pi]$ because $F(\cdot,t)$ is, and $G(\cdot,0)$ is proper, but $G(\cdot,\pi/2)$ is not proper because it is given by the product of $\arctan$ functions.

  • $\begingroup$ You answered my question, however is the result still false when $m>n$ where $n = \dim M$? $\endgroup$ – abcdef Apr 30 '18 at 9:10
  • $\begingroup$ @abcdef It's still false when $m > n$. I added an example in my answer. Also, I edited my answer so that the counterexample is analytic rather than merely $C^\infty$. $\endgroup$ – Matthew Kvalheim Apr 30 '18 at 9:25
  • $\begingroup$ Thank you! Sorry my trivial last question, I was focused on other kinds of counterexamples. $\endgroup$ – abcdef Apr 30 '18 at 9:29
  • $\begingroup$ @abcdef no problem. Also, I removed my comment about Hirsch's book. I was being silly: if $M$ is compact, every map with domain $M$ is automatically proper :-). $\endgroup$ – Matthew Kvalheim Apr 30 '18 at 9:31

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